Leetcode 67. Add Binary大数相加

Given two binary strings, return their sum (also a binary string).

The input strings are both non-empty and contains only characters 1 or 0.

Example 1:

Input: a = "11", b = "1"
Output: "100"

Example 2:

Input: a = "1010", b = "1011"
Output: "10101"

题目链接：https://leetcode.com/problems/add-binary/

思路：之前做过大数相乘，这把遇到了大数相加，虽然是二进制版本的，但原理一样

class Solution {
public:
    string addBinary(string a, string b) {
        int len1=a.length(),len2=b.length();
        string res="";
        int t=0,c=0,n=min(len1,len2);
        for(int i=1;i<=n;i++)
        {
            t=(c+a[len1-i]+b[len2-i]-96);
            c=t/2;
            res+=(t%2+48);
        }
        if(len1>=len2)
        {
            for(int i=len1-n-1;i>=0;i--)
            {
                t=(c+a[i]-48);
                c=t/2;
                res+=t%2+48;
            }
        }
        else
        {
            for(int i=len2-n-1;i>=0;i--)
            {
                t=(c+b[i]-48);
                c=t/2;
                res+=t%2+48;
            }
        }
        if(c)
            res+="1";
        char ch;
        int len=res.length();
        for(int i=0;i<len/2;i++)
        {
            ch=res[i];
            res[i]=res[len-1-i];
            res[len-1-i]=ch;
        }
        return res;
    }
};

也看了一下网上比较短的代码，可以学习一下

class Solution {
public:
    string addBinary(string a, string b) {
        string re;
        int c=0;
        int i=a.size()-1,j=b.size()-1;
        while(i>=0||j>=0||c!=0){
            int f = i>=0?a[i]-'0':0;
            int s = j>=0?b[j]-'0':0;
            int cur = f+s+c;
            if(cur>1)
                c=1;
            else
                c=0;
            if(cur==0||cur==2)
                re="0"+re;
            else
                re="1"+re;
            i--,j--;
        }
        return re;
    }
};

参考链接：https://leetcode.com/problems/add-binary/discuss/219381/Short-and-Fast-C%2B%2B-Sol